An equilateral triangle and an isosceles triangle share a common side. what is the measure of angle abc? i need with question 13 asap!

The measure of the angle ∠ABC is 108°

Step-by-step explanation:

Find the measure of the vertex angle ∠ABD of an isosceles triangle

we know that

An isosceles triangle has two equal sides and two equal angles

In this problem

∠BDA=∠BAD= 66° > the angles of the base are equals

Find the measure of the vertex angle

∠ABD = 180° - 2×66° = 48° (the sum of the internal angles of a triangle is  

                                             equal to 180°)

then,

Find the measure of the angle  ∠CBD in the equilateral triangle

we know that

A equilateral triangle has three  equal sides and three equal angles

The measure of the internal angle in a equilateral triangle is 60°

so

∠CBD= 60°

Find the measure of the angle ∠ABC

∠ABC=∠ABD+∠DBC

substitute the values

∠ABC= 48°+60° = 108°

Therefore

The answer is 108°

The measure of the angle ∠ABC is 108°

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Step 1

Find the measure of the vertex angle ∠ABD of an isosceles triangle

we know that

An isosceles triangle has two equal sides and two equal angles

In this problem

∠ABD=∠BAD= ----> the angles of the base are equals

Find the measure of the vertex angle

∠ABD= ------> the sum of the internal angles of a triangle is equal to

Step 2

Find the measure of the angle  ∠CBD in the equilateral triangle

we know that

A equilateral triangle has three  equal sides and three equal angles

The measure of the internal angle in a equilateral triangle is

so

∠CBD=

Step 3

Find the measure of the angle ∠ABC

∠ABC=∠ABD+∠DBC

substitute the values

∠ABC=

therefore

the answer is

the measure of the angle ∠ABC is

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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